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This article mainly shows you how java solves the closest sum of three problems. The content is simple and easy to understand. It is clearly organized. I hope it can help you solve your doubts. Let Xiaobian take you to study and learn about "java how to solve the closest sum of three problems."

Given an array of n integers nums and a target value target. Find three integers in nums whose sum is closest to target. Return the sum of these three numbers. Assume that there is only one answer for each set of inputs.

For example, given the array nums = [-1,2,1,-4], and target = 1. The sum of the three numbers closest to target is 2. (-1 + 2 + 1 = 2).

Analysis:

The second line sorts the array from smallest to largest, and then the fifth line has num[i]!= num[i-1], he is to prevent duplication, for example, sorted array [-8,-1,-1, 2,5] then he will start from the first-1 to find [-1,-1, 2] the sum of this group is 0, if there is no code above, then he can also find [-1,-1, 2] from the second-1, so repeated. Then the following is relatively simple, the sum of three numbers minus one of them, the rest is the sum of the other two numbers, because it is ordered, so here is from both ends to the middle, if found directly into the mList, and then look for, if less than sum, indicating that the sum of two numbers is small, so lo++, because here is ordered, from both ends to the middle, so lo can only be added can not be subtracted, hi can only be subtracted can not be added. When the sum of two numbers is greater than sum, hi--.

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